Business Applications — Question 7

Question 7

A company’s profit $P(x)$ (in dollars) depends on the number of items $x$ produced per week: $$P(x) = -5x^2 + 300x - 2000$$

• Determine the production level $x$ that maximizes profit.

• Compute the maximum profit.

• Find the production levels $x$ at which the profit is zero.

Question 7 - Solution

(a) Maximize Profit:

$$P'(x) = \frac{d}{dx}(-5x^2 + 300x - 2000) = -10x + 300$$

Set $P'(x) = 0$: $$-10x + 300 = 0 \quad \Rightarrow \quad x = 30$$

Second derivative: $$P''(x) = -10 < 0$$ Hence, profit is maximized at $x = 30$.

(b) Maximum Profit:

$$P(30) = -5(30)^2 + 300(30) - 2000 = -4500 + 9000 - 2000 = 2500$$

(c) Break-even Points (Profit = 0):

$$-5x^2 + 300x - 2000 = 0$$

Divide through by 5: $$x^2 - 60x + 400 = 0$$

Solve using the quadratic formula: $$x = \frac{60 \pm \sqrt{3600 - 1600}}{2} = \frac{60 \pm \sqrt{2000}}{2} = 30 \pm 10\sqrt{5}$$

$$\boxed{ \begin{aligned} &\text{Maximum profit occurs at } x = 30 \text{ units, } P_{\max} = 2500, \\ &\text{Profit is zero at } x = 30 - 10\sqrt{5}, \; 30 + 10\sqrt{5}. \end{aligned} }$$